We have two wire `A` and `B` of the same mass and the same material. The diameter of the wire `A` is half of that `B`. If the resistance of wire `A` is `24 ohm` them the resistance of wire `B` will be

Options

A. `3 Omega`
B. `1.5 Omega`
C. `4.5 Omega`
D. `6 Omega`

Answer

B

Explanation

`r_A = (1)/(2) r_B`
`m = d pi r_A^2 l_A = d pi r_B^2 l_B`
`(l_A)/(l_B) = ((r_B)/(r_A))^2 = 4`
`(R_B)/(R_A) = (rho l_B// pi r_B^2)/(rho l_A//pi r_A^2) = ((l_B)/(l_A))((r_A)/(r_B))^2`
`(R_B)/(24) = ((1)/(4))((1)/(2))^2 = (1)/(16)`
`R_B = 1.5 Omega`.